극대와 극소

$f\,'(0) = 0$

$f\,'(0)$ is not defined

When $x=0$, the curve $y=f(x)$ appears to have a flat tangent line, so the $x=0$ is a critical point. However, it is not a local extremum: it is not true that $f(0) \geq f(x)$ for all $x$ near 0, and it is not true that $f(0) \leq f(x)$ for all $x$ near 0.

To the right of the $x$-axis, there is a spike where the derivative of $f(x)$ does not exist. The $x$-value corresponding to this spike (call it $a$) is a singular point, and $f(x)$ has a local maximum at $x=a$.

The $x$-coordinate corresponding to the blue dot (let's call it $a$) is a critical point, because the tangent line to $f(x)$ at $x=a$ is horizontal. There is no lower point nearby, and actually no lower point on the whole interval shown, so $f(x)$ has both a local minimum and a global minimum at $x=a$.

If a function is not continuous at a point, then it is not differentiable at that point. So, the $x$-coordinate corresponding to the discontinuity (let's call it $b$) is a singular point. Values of $f(x)$ immediately to the right of $b$ are lower, and values immediately to the left of $b$ are higher, so $f(x)$ has no local (or global) extremum at $x=b$.

One possible answer is shown below.

For every $x$ in the red interval shown below, $f(2) \geq f(x)$, so $f(2)$ is a local maximum. However, the point marked with a blue dot shows that $f(x)>f(2)$ for some $x$, so $f(2)$ is not a global maximum.

For the first curve, the function's value at $x=2$ (that is, the $y$-value of the solid dot) is higher than anything around it. So, it's a local maximum.

For the second curve, the function's value at $x=2$ (that is, the $y$-value of the solid dot) is higher than everything to the left, but lower than values immediately to the right. (On the graph reproduced below, $f(x)$ is higher than everything in the red section, and lower than everything in the blue section.) So, it is neither a local max nor a local min.

Similarly, for the third curve, $f(2)$ is lower than the values to the right of it, and higher than values to the left of it, so it is neither a local minimum nor a local maximum.

In the final curve, $f(2)$ (remember–this is the $y$-value of the solid dot) is higher than everything immediately to the left or right of it (for instance, over the interval marked in red below), so it is a local maximum.

(a) $g$ has neither a local maximum nor local minimum at $x = 2$ because $f'$ does not change sign at $x = 2$. (b) Because $g'(2) = 0$ and $g'$ is positive for $1 < x < 2$ and for $2 < x < 3$, it must be that $g'$ is decreasing for $1 < x < 2$ and $g'$ is increasing for $2 < x < 3$. Since $g'$ changes from decreasing to increasing at $x = 2$, it follows that $g'$ has a local minimum at $x = 2$ and thus $g''(2) = 0$. In addition, $g''(x)$ must change sign from negative to positive at $x = 2$. In particular, $g''$ is negative for $1 < x < 2$ and positive for $2 < x < 3$. (c) $g$ has a horizontal tangent line at $x = 2$, is increasing before $x = 2$ and decreasing after $x = 2$, and concave down before $x = 2$ and concave up after $x = 2$, and thus $g$ has a point of inflection at $x = 2$.

(a) Note that $p''(x)=0$ at $x=-1$ and $x=2$. The second derivative sign chart for $p$ shows that $p''(x)$ is negative for $-1 < x < 2$ and positive for all other values of $x$. Since $p''$ changes sign at $x = -1$ and $x = 2$, $p$ has points of inflection at $x = -1$ and $x = 2$. In particular, $p$ is concave down for $-1 < x < 2$ and concave up for all other values of $x$. (b) If we compute $p''\left(\frac{\sqrt{5}-1}{2} \right)$, we find that $p''\left(\frac{\sqrt{5}-1}{2} \right) \approx -1.205$, which means that $p$ is concave down at $x = \frac{\sqrt{5}-1}{2}$. Since $p$ has a horizontal tangent line at this point, it follows that $p$ has a local maximum at $x = \frac{\sqrt{5}-1}{2}$ by the second derivative test. (c) Since $p(2) = \frac{12}{e^2}$ and $p'(2) = -\frac{5}{e^2}$, the equation of the tangent line to $p$ at $x = 2$ is given by $y = -\frac{5}{e^2}(x-2) + \frac{12}{e^2}$. To determine whether the tangent line lies above or below the graph of $y=p(x)$ at $x = 2$, we consider the concavity of $p$. Recall we know that $p''(2) = 0$ and that $p$ is concave down to the left of $x = 2$ and concave up to the right. Thus, the tangent line lies neither solely below nor solely above the graph of $y = p(x)$ at $x = 2$; it lies below $y = p(x)$ to the right of $x = 2$ and above $y = p(x)$ to the left of $x = 2$.